1. System Maintainability Modeling & Analysis Leadership in Engineering
Systems Engineering Program
Department of Engineering Management, Information and Systems
EMIS 7305/5305
Systems Reliability, Supportability and Availability Analysis
System Maintainability Modeling & Analysis
Dr. Jerrell T. Stracener, SAE Fellow
Leadership in Engineering

2. Why Do Maintainability Modeling and Analysis?
To identify the important issues
To quantify and prioritize these issues
To build better design and support systems 2

3. Bottoms Up Models
Provide output to monitor design progress vs. requirements
Provide input data for life cycle cost
Provide trade-off capability
Design features vs. maintainability requirements
Performance vs. maintainability requirements
Provide Justification for maintenance improvements perceived as the design progresses 3

4. Bottoms Up Models
Provide the basis for maintainability guarantees/demonstration
Provide inputs to warranty requirements
Provide maintenance data for the logistic support analysis record
Support post delivery design changes
Inputs
Task Time (MH)
Task Frequency (MTBM)
Number of Personnel-Elapsed Time (hours)
For each repairable item 4

5. Bottoms Up Models Input Data Sources Task Frequency
Reliability predictions de-rated to account for non-relevant failures
Because many failures are repaired on equipment, the off equipment task frequency will be less than the task frequency for on equipment 5

6. Bottoms Up Models Input Data Sources (Continued) Task Time
Touch time vs. total time
That time expended by the technician to effect the repair
Touch time is design controllable
Total Time
Includes the time that the technician expends in “Overhead” functions such as part procurement and paper work
Are developed from industrial engineering data and analyst’s estimates 6

7. Task Analysis Model Task analysis modeling estimates repair time
MIL-HDK-472 method V
Spreadsheet template
Allow parallel and multi-person tasks estimation
Calculates elapsed time and staff hours
Reports each task element and total repair time
Sums staff hours by repairmen type
Estimates impact of hard to reach/see tasks 7

8. When considering probability distributions in general, the time dependency between probability of repair and the time allocated for repair can be expected to produce one of the following probability distribution functions:
Normal – Applies to relatively straightforward maintenance tasks and repair actions that consistently require a fixed amount of time to complete with little variation
Exponential – Applies to maintenance tasks involving part substitution methods of failure isolation in large systems that result in a constant repair rate.
Lognormal - Applies to Most maintenance tasks and repair actions comprised of several subsidiary tasks of unequal frequency and time duration.

9. The Exponential Model:
Definition
A random variable X is said to have the Exponential
Distribution with parameters , where  > 0, if the
probability density function of X is:
, for x  0
, elsewhere

10. Properties of the Exponential Model:
Probability Distribution Function
for x < 0
for x  0
Note: the Exponential Distribution is said to be
without memory, i.e.
P(X > x1 + x2 | X > x1) = P(X > x2)

11. Properties of the Exponential Model:
Mean or Expected Value
Standard Deviation

12. Normal Distribution:
A random variable X is said to have a normal (or
Gaussian) distribution with parameters  and ,
where -  <  <  and  > 0, with probability
density function
-  < x < 
where  = … and e =
f(x) x

13. Normal Distribution:
Mean or expected value of X
Mean = E(X) = 
Median value of X
X0.5 = 
Standard deviation

14. Normal Distribution:
Standard Normal Distribution
If X ~ N(, ) and if , then Z ~ N(0, 1).
A normal distribution with  = 0 and  = 1, is called
the standard normal distribution.

15.  x z
Normal Distribution:
f(x)
f(z)

16. Standard Normal Distribution Table of Probabilities
Enter table with
and find the
value of  z  z
f(z)

17. Normal Distribution - Example:
The time it takes a field engineer to restore a
function in a logistics system can be modeled with
a normal distribution having mean value 1.25 hours
and standard deviation 0.46 hours. What is the
probability that the time is between 1.00 and 1.75
hours? If we view 2 hours as a critically time,
what is the probability that actual time to restore
the function will exceed this value?

18. Normal Distribution - Example Solution:

19. Normal Distribution - Example Solution:

20. The Lognormal Model:
Definition - A random variable X is said to have the
Lognormal Distribution with parameters  and ,
where  > 0 and  > 0, if the probability density
function of X is:
, for x > 0
, for x  0

21. Properties of the Lognormal Distribution
Probability Distribution Function
where (z) is the cumulative probability distribution
function of N(0,1)
Rule: If T ~ LN(,), then Y = lnT ~ N(,)

22. Properties of the Lognormal Model:
Mean or Expected Value
Variance
Median

23. Lognormal Model example
The elapsed time (hours) to repair an item is a
random variable. Based on analysis of data, elapsed
time to repair can be modeled by a lognormal
distribution with parameters  = 0.25 and  = 0.50.
a. What is the probability that an elapsed time to
repair will exceed 0.50 hours?
b. What is the probability that an elapsed time to
repair will be less than 1.2 hours?
c. What is the median elapsed time to repair?
d. What is the probability that an elapsed time to
repair will exceed the mean elapsed time to repair?
e. Sketch the cumulative probability distribution
function.

24. Lognormal Model example - solution
a. What is the probability that an elapsed time to
repair will exceed 0.50 hours?
X ~ LN(, ) where  = 0.25 and  = 0.50
note that:
Y = lnX ~ N(, )
P(X > 0.50) = P(lnX > )

25. Lognormal Model example
b. What is the probability that an elapsed time to
repair will be less than 1.2 hours?
P(X < 1.20) = P(lnX < ln1.20)

26. Lognormal Model example
c. What is the median elapsed time to repair?
P(X < x0.5) = 0.5
therefore

27. Lognormal Model example
d. What is the probability that an elapsed time to
repair will exceed the mean elapsed time to repair?

28. Lognormal Model example
P(X > MTTR) = P(X > 1.455)
= P(lnX > 0.375)

29. Lognormal Model example
e. Sketch the cumulative probability distribution
function.
tmax

30. 95th Percentile / MTTR Ratio
If repair time, T, has a lognormal distribution with parameters μ and σ, then
95th percentile of time to repair
Mean Time To Repair
Ratio of 95th percentile to mean time to repair

31. 95th Percentile / MTTR Ratio

32. 95th Percentile / MTTR Ratio
Since
Then
and

33. Analysis of Combination of Repair Times

34. Active Maintenance Commences Faulty Item Identified
Corrective Maintenance Cycle
Failure Occurs
Detection
Failure Confirmed
Preparation for
Maintenance
Active Maintenance Commences
Location and
Isolation
Faulty Item Identified
Disassembly
(Access)
Disassembly Complete or
Removal of
Fault Item
Repair of
Equipment
Installation of
Spare/Repair Part
Re-assembly
Re-assembly Complete
Alignment
and Adjustment
Condition
Verification 34
Repair Completed

35. If X1, X2, ..., Xn are independent random variables
with means 1, 2, ..., n and variances 12, 22, ...,
n2, respectively, and if a1, a2, … an are real numbers
then the random variable
has mean
and variance
Linear Combinations of Random Variables

36. where a1, a2, … an are real numbers
Linear Combinations of Random Variables
If X1, X2, ..., Xn are independent random variables
having Normal Distributions with means 1, 2, ..., n
and variances 12, 22, ..., n2, respectively, and if
then
where
and
where a1, a2, … an are real numbers

37. Example

38. Generating Random Samples using Monte Carlo Simulation38

39. Population vs. Sample Population
the total of all possible values (measurement,
counts, etc.) of a particular characteristic for a
specific group of objects.
Sample
a part of a population selected according to some
rule or plan.
Why sample?
- Population does not exist
- Sampling and testing is destructive

40. Sampling Characteristics that distinguish one type of sample
from another:
the manner in which the sample was obtained
the purpose for which the sample was obtained

41. Types of Samples Simple Random Sample
The sample X1, X2, ... ,Xn is a random sample if
X1, X2, ... , Xn are independent identically
distributed random variables.
Remark: Each value in the population has an
equal and independent chance of being included
in the sample.
Stratified Random Sample
The population is first subdivided into
sub-populations for strata, and a simple random
sample is drawn from each strata

42. Types of Samples (continued)
Censored Samples
Type I Censoring - Sample is terminated at a
fixed time, t0. The sample consists of K times to
failure plus the information that n-k items
survived the fixed time of truncation.
Type II Censoring - Sampling is terminated
upon the Kth failure. The sample consists of K
times to failure, plus information that n-k items
survived the random time of truncation, tk.
Progressive Censoring - Sampling is reduced in
stage.

43. Uniform Probability Integral Transformation
For any random variable Y with probability density
function f(y), the variable
is uniformly distributed over (0, 1), or F(y) has the
probability density function

44. Uniform Probability Integral Transformation
Remark: the cumulative probability distribution
function for any continuous random variable is
uniformly distributed over the interval (0, 1).

45. Generating Random Numbers
f(y)
F(y) y
1.0
0.8
0.6
0.4
0.2 ri yi

46. Generating Random Numbers
Generating values of a random variable using the
probability integral transformation to generate a
random value y from a given probability density
function f(y):
1. Generate a random value rU from a uniform
distribution over (0, 1).
2. Set rU = F(y)
3. Solve the resulting expression for y.

47. Generating Random Numbers with Excel
From the Tools menu, look for Data Analysis.

48. Generating Random Numbers with Excel
If it is not there, you must install it.
Generating Random Numbers with Excel

49. Generating Random Numbers with Excel
Once you select Data Analysis, the following window will appear. Scroll down to “Random Number Generation” and select it, then press “OK”

50. Generating Random Numbers with Excel
Choose which distribution you would like. Use uniform for an exponential or weibull distribution or normal for a normal or lognormal distribution

51. Generating Random Numbers with Excel
Uniform Distribution, U(0, 1).
Select “Uniform” under the “Distribution” menu.
Type in “1” for number of variables and 10 for number of random numbers. Then press OK random numbers of uniform distribution will now appear on a new chart.

52. Generating Random Numbers with Excel
Normal Distribution, N(μ, σ).
Select “Normal” under the “Distribution” menu.
Type in “1” for number of variables and 10 for number of random numbers. Enter the values for the mean (m) and standard deviation (s) then press OK. 10 random numbers of uniform distribution will now appear on a new chart.

53. Generating Random Values from an Exponential
Distribution E() with Excel
First generate n random variables, r1, r2, …, rn, from
U(0, 1).
Select “Uniform” under the “Distribution” menu.
Type in “1” for number of variables and 10 for number of random numbers. Then press OK random numbers of uniform distribution will now appear on a new chart.

54. Generating Random Values from an Exponential
Distribution E() with Excel
Select a θ that you would like to use, we will use θ = 5.
Type in the equation xi= -ln(1 - ri), with filling in θ as 5, and ri as cell A1 (=-5*LN(1-A1)). Now with that cell selected, place the cursor over the bottom right hand corner of the cell. A cross will appear, drag this cross down to B10. This will transfer that equation to the cells below. Now we have n random values from the exponential distribution with parameter θ=5 in cells B1 - B10.

55. Generating Random Values from an Weibull
Distribution W(β, ) with Excel
First generate n random variables, r1, r2, …, rn, from U(0, 1).
Select “Uniform” under the “Distribution” menu.
Type in “1” for number of variables and 10 for number of random numbers. Then press OK random numbers of uniform distribution will now appear on a new chart.

56. Generating Random Values from an Weibull
Distribution W(β, ) with Excel
Select a β and θ that you would like to use, we will use β =20, θ =
Type in the equation xi = [-ln(1 - ri)]1/, with filling in β as 20, θ as 100, and ri as cell A1 (=100*(-LN(1-A1))^(1/20)). Now transfer that equation to the cells below. Now we have n random variables from the Weibull distribution with parameters β =20 and θ =100 in cells B1 - B10.

57. Generating Random Values from an Lognormal
Distribution LN(μ, σ) with Excel
First generate n random variables, r1, r2, …, rn, from N(0, 1).
Select “Normal” under the “Distribution” menu.
Type in “1” for number of variables and 10 for number of random numbers. Enter 0 for the mean and 1 for standard deviation then press OK random numbers of uniform distribution will now appear on a new chart.

58. Generating Random Values from an Lognormal
Distribution LN(μ, σ) with Excel
Select a μ and s that you would like to use, we will use μ = 2, σ = 1.
Type in the equation , with filling in μ as 2, σ as 1, and ri as cell A1 (=EXP(2+A1*1)). Now transfer that equation to the cells below. Now we have an Lognormal distribution in cells B1 - B10.

59. Flow Chart of Monte Carlo Simulation method
Input 1: Statistical distribution
for each component variable.
Input 2: Relationship
between component
variables and system
performance
Select a random value from
each of these distributions
Calculate the value of system
performance for a system
composed of components with the
values obtained in the previous step.
Output: Summarize and plot resulting
values of system performance. This
provides an approximation of the
distribution of system performance.
Repeat n
times

60. Sample and Size Error Bands
Because Monte Carlo simulation involves randomly
selected values, the results are subject to statistical
fluctuations.
Any estimate will not be exact but will have an
associated error band.
The larger the number of trials in the simulation,
the more precise the final results.
We can obtain as small an error as is desired by
conducting sufficient trials
In practice, the allowable error is generally specified,
and this information is used to determine the required trials

61. Drawbacks of the Monte Carlo Simulation
there is frequently no way of determining
whether any of the variables are dominant or
more important than others without making repeated
simulations
if a change is made in one variable, the entire
simulation must be redone
the method may require developing a
complex computer program
if a large number of trials are required, a great
deal of computer time may be needed to obtain
the necessary results

62. System Mean Time to Repair, MTTRS
System without redundancy E1 E2 En 62

63. Systems Maintainability Analysis Examples
Example 1: Compute the mean time to repair at the system level for the following system.
Solution:
MTTF = 500 h
MTTR = h
MTTF = 400 h
MTTR = 2.5 h
MTTF = 250 h
MTTR = h
MTTF = 100 h
MTTR = 0.5 h 63

64. Maintainability Prediction
Example 2: How does the MTTRs of the system in the previous example change if an active redundancy is introduced to the element with MTTF = 100h?
MTTF = 100 h
MTTR = h
MTTR = 0.5 h
MTTF = 500 h
MTTR = h
MTTF = 400 h
MTTR = 2.5 h
MTTF = 250 h
MTTR = h
Solution: 64

65. Maintainability Math Task Time : MTTR ~ N(,σ)|LN(,σ)
Task frequency : MTBM ~ E()
Crew Size : CS is constant
MMH/FH = CS*MTTR/MTBM
T95%=
N : +1.645σ
LN : e+1.645σ
T50%=
N : 
LN : e
MTTR=
LN : e+½σ^2 65

66. Maintainability Math: MTBM
Given that MTBM is exponential
MTBM = inverse of sum of inverse MTBMs
Allocation
Given a top level MTBM and a complexity factor for components Ci such that ΣCi =1.
MTBMi=MTBM*Ci
Roll-up
Given components MTBMi
MTBM=1/Σ(1/MTBMi) 66

67. Maintainability Math: MMH/FH
MMH/FH = sum of MMH/FHs
Allocation
Given a top level MMH/FH and a complexity factor for component repairs Ci such that ΣCi =1.
MMH/FHi=MFH/FH*Ci
Roll-up
Given component’s MMH/FHi or MTTRi , CSi and MTBMi
MMH/FH=ΣMMH/FHi =ΣCSi*MTTRi/MTBMi 67

68. Maintainability Math: MTTR
MTTR = weighted sum of MTTRs
Weighting factor is frequency of maintenance = 1/MTBM
Roll-up
Given components MTTRi and MTBMi
MTTR = MTBM*Σ(MTTRi /MTBMi)
= (Σ(MTTRi /MTBMi)/Σ(1/MTBMi) 68

69. Maintainability Math: CS
CS is computed from MMH/FM, MTBM and MTTR
CS=MMH/FH*MTBM/MTTR
Roll-up
Given components MTTRi , CSi and MTBMi
CSaverage= MMH/FH*MTBM/MTTR
=(ΣCSi*MTTRi/MTBMi)/Σ(MTTRi /MTBMi) 69

70. Maintainability Math: MTBM
Given that MTBM is exponential
MTBM = inverse of sum of inverse MTBMs
Allocation
Given a top level MTBM and a complexity factor for components Ci such that ΣCi =1.
MTBMi=MTBM*Ci
Roll-up
Given components MTBMi
MTBM=1/Σ(1/MTBMi) 70

71. Maintainability Math: MMH/FH
MMH/FH = sum of MMH/FHs
Allocation
Given a top level MMH/FH and a complexity factor for component repairs Ci such that ΣCi =1.
MMH/FHi=MFH/FH*Ci
Roll-up
Given component’s MMH/FHi or MTTRi , CSi and MTBMi
MMH/FH=ΣMMH/FHi =ΣCSi*MTTRi/MTBMi 71

72. Maintainability Math: MTTR
MTTR = weighted sum of MTTRs
Weighting factor is frequency of maintenance = 1/MTBM
Roll-up
Given components MTTRi and MTBMi
MTTR = MTBM*Σ(MTTRi /MTBMi)
= (Σ(MTTRi /MTBMi)/Σ(1/MTBMi) 72

73. Example: Roll-up
Given the following R&M characteristics for the items comprising a subsystem, what are the subsystem R&M characteristics? 73

74. Solution: Roll-up MTBM =1/(Σ1/MTBMi) MMH/FH =ΣMMH/FHi
MTTR =(ΣMTTRi/MTBMi)/(Σ1/MTBMi)
CS =MMH/FH*MTBM/MTTR
A B C D E F G H 1 2 3 4 5 74

75. Analysis and Model Selection
Maintainability Data
Analysis and Model Selection 75

76. Estimation of the Mean - Normal Distribution
X1, X2, …, Xn is a random sample of size n from
N(, ), where both µ & σ are unknown.
Point Estimate of 
Point Estimate of s ^ 76

77. Estimation of Lognormal Distribution
Random sample of size n, X1, X2, ... , Xn from
LN (, )
Let Yi = ln Xi for i = 1, 2, ..., n
Treat Y1, Y2, ... , Yn as a random sample from
N(, )
Estimate  and  using the Normal Distribution
Methods 77

78. Estimation of the Mean of a Lognormal Distribution
Mean or Expected value of
Point Estimate of mean
where and are point estimates of and respectively. ^ ^ 78

79. 95th Percentile / MTTR Ratio
If repair time, T, has a lognormal distribution with parameters μ and σ, then
95th percentile of time to repair
Mean Time To Repair
Ratio of 95th percentile to mean time to repair 79

80. Procedure for Prediction of 95th percentile time
to repair using the predicted MTTR
Obtain a random sample of n times to repair for a given subsystem, t1, t2, …, tn
Utilize probability plotting on lognormal probability paper and/or use a statistical goodness of fit test to test the validity of the lognormal distribution
Assuming the results indicate that the lognormal distribution provides a “good” fit to the data, estimate σ as follows: 80

81. Predict the 95th percentile repair time as follows:
Procedure for Prediction of 95th percentile time to repair using the predicted MTTR
Estimate r as follows:
Predict the 95th percentile repair time as follows: 81